picture: http://postimg.org/image/v0dkl32id/
In the figure, an isosceles right-angled triangle X1Y1Z1 is inscribed in a circle C1 and C2 is a circle inscribed in triangle X1Y1Z1. Another isosceles right-angled
triangle X2Y2Z2 is inscribed in C2 and C3 is a circle inscribed in triangle X2Y2Z2.
The process is repeated indefinitely. It is given that the radius of C1 is 10cm.
a) show that the radius of C2 is 10[(√2)-1] cm
b) show that the radius of C3 is 10(3-2√2) cm
I can just find out X1Y1=20 cm and X1Z1=10√2 cm
I think it shoud be related to in-centre, circumcentre or tangents from external
points, but I still do not know how to calculate.
Please help, thank you!!
F.5 Maths geometry
2015-04-07 11:25 pm
回答 (2)
2015-04-08 2:22 am
✔ 最佳答案
(a)Since ∠X1Z1Y1 = 90°,then X1Y1 is the diameter of C1 (∠ insemi-circle is rt.∠)
X1Y1 = 10 × 2 cm = 20 cm
As ΔX1Y1Z1 is isosceles, let X1Z1= Y1Z1 = y cm
X1Y1² = X1Z1² + Y1Z1²
20² = y² + y²
2y² = 400
y = 10√2
X1Z1 = Y1Z1 = 10√2
Let X1Y1, Y1Z1 and X1Z1touches C1 at P1, Q1 and R1respectively.
Let X1P1 = X1R1 = a cm, Y1P1= Y1Q1 = b cm, Z1Q1 = Z1R1= c cm
X1Y1 : a + b = 20 ... [1]
Y1Z1 : b + c = 10√2 ... [2]
X1Z1 : a + c = 10√2 ... [3]
[1] + [2] + [3] :
2a + 2b + 2c = 20 + 20√2
a + b + c = 10 + 10√2 ... [4]
[4] - [1] :
c = 10(√2) - 10
c = 10 [(√2) - 1]
Radius of C2 = c cm
Radius of C2 = 10 [(√2) - 1] cm
(b)
(Radius of C1) : (Radius of C2) = 10 : 10 [(√2) - 1]
(Radius of C1) : (Radius of C2) = 1 : [(√2) - 1]
Similarly, (Radius of C2) : (Radius of C3) =1 : [(√2) -1]
{10[(√2) - 1] cm} : (Radius of C3) = 1 : [(√2) - 1]
Radius of C3
= 10 [(√2) - 1]² cm
= 10 [2 - 2(√2) + 1] cm
= 10 (3 - 2√2) cm
2015-04-07 18:29:01 補充：
(a) Alternative method :
Use the formula : r = A/s
where :
r : the radius of inscribed circle of a right-angled triangle
A : the area of the right-angled triangle
s : half of the perimeter of the right-angle triangle
2015-04-07 18:29:46 補充：
A = (1/2) × 10(√2) × 10(√2) = 100 cm²
s = [20 + 10(√2) + 10√2] / 2 cm = 10[(√2) + 1] cm
r
= 100 / 10[(√2) + 1] cm
= 10[(√2) - 1] / [(√2) + 1][(√2) - 1] cm
= 10[(√2) - 1] cm
2015-04-08 2:26 am
a) Let O be center of circle C2, so the perpendicular distance from O to the 3 sides of triangle X1Y1Z1 is the radius of C2, let it be r.
Since X1Y1 = X2Y2 = 10 sqrt 2 as calculated.
So area of triangle X1Y1Z1 = (10 sqrt 2)(10 sqrt 2)/2 = 100.
But area of triangle X1Y1Z1 = area of traingle OX1Y1 + area of triangle OX1Z1 + area of triangle OY1Z1
= r(10 sqrt 2)/2 + r(10 sqrt 2)/2 + r(20)/2 = 10r sqrt 2 + 10r = 10r(sqrt 2 + 1)
That is 10r(sqrt 2 + 1) = 100
r = 10/(sqrt 2 + 1)
= 10(sqrt 2 - 1)/[(sqrt 2 + 1)(sqrt 2 -1)]
= 10 (sqrt 2 - 1).
b)
Pattern X1Y1Z1C2 is similar to pattern X2Y2Z2C3.
radius of C3/radius of C2 = X2Y2/X1Y1
radius of C3/10(sqrt2 - 1) = 20(sqrt 2 - 1)/20
so radius of C3 = 10(sqrt 2 - 1)^2 = 10(2 + 1 - 2sqrt 2)
= 10(3 - 2 sqrt 2)
Since X1Y1 = X2Y2 = 10 sqrt 2 as calculated.
So area of triangle X1Y1Z1 = (10 sqrt 2)(10 sqrt 2)/2 = 100.
But area of triangle X1Y1Z1 = area of traingle OX1Y1 + area of triangle OX1Z1 + area of triangle OY1Z1
= r(10 sqrt 2)/2 + r(10 sqrt 2)/2 + r(20)/2 = 10r sqrt 2 + 10r = 10r(sqrt 2 + 1)
That is 10r(sqrt 2 + 1) = 100
r = 10/(sqrt 2 + 1)
= 10(sqrt 2 - 1)/[(sqrt 2 + 1)(sqrt 2 -1)]
= 10 (sqrt 2 - 1).
b)
Pattern X1Y1Z1C2 is similar to pattern X2Y2Z2C3.
radius of C3/radius of C2 = X2Y2/X1Y1
radius of C3/10(sqrt2 - 1) = 20(sqrt 2 - 1)/20
so radius of C3 = 10(sqrt 2 - 1)^2 = 10(2 + 1 - 2sqrt 2)
= 10(3 - 2 sqrt 2)
收錄日期: 2021-04-15 18:58:24
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